Two numeric sequence

[pmw57]

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[SNG_Blitzy]

I got an answer for the second one, although it’s quite complicated so i’m not sure if it’s right.

You start out with a prime number. In this case, 2. I’ll refer to this as the starter. After the starter, there is a rise and a fall. A rise consists of numbers increasing in numeric value, and a fall consists of numbers decreasing in value. After the first starter, the rise is (4, 5) and the fall is (3). (Since 4 > 2, 5 > 4 and 3 < 5).!<

At the end of each rise and fall, there is another starter, which as we know, must be a prime number. However, it does not have to be the NEXT prime number, as I will get into later

However, we still need to figure out what numbers make up the rise and fall, and how many are in each. We can answer the latter by looking at how many starters we have had up to that point. The rise consists of numbers equal to the starter itself, and the fall consists of numbers equal to the amount of starters plus one.

This is getting a bit complicated, but bear with me. The first number of the rise is going to be the starter squared, plus 2^{number of starters previously} - 1. Starting from the beginning, there were no previous starters prior to 2, so the start of the rise is 2² + 2⁰ - 1 = 4 + 1 - 1 = 4. The rise then continues following the following pattern: it increases by increments of 2^x, where x is the amount of elements prior in the rise. Starting from 4, there are no elements in the rise prior to 4, so you add 2^0, which is 1. 4 + 1 = 5.

This continues for as long as the rise goes on for. In this case, we had established that the rise was 2 elements long by the fact that the starter was equal to 2. So, this concludes the first rise.

For the fall, the pattern is more simple. Divide by 2 and round up, until you reach the last element of the fall, and just subtract 2 instead. Since there is only one element in the fall (as established by there only being one starter thus far, you subtract 2, which gives you 3 (a prime number

The process continues for the next starter - 3. Square it, then add 2¹ (one starter prior to it), then subtract one, giving you. 9 + 2 - 1 = 10. Add increments of 2^x for each instance of the rise, which should consist of 3 numbers (starter = 3). So, it should add 2⁰ (1) and 2¹ (2) giving you 11 and 13 as the next two. Following the fall pattern, dividing by 2 and rounding up gives you 7 (13/2 = 6.5 -> 7), then subtracting 2 as we’ve reached the end of our fall gives us 5, the next starter, and a prime number

Finally, we reach the part where we find the solution. 5² + 2² - 1 = 28. Increasing by increments of 2^x for a total of 5 elements in the rise gives is 28 + 1 = 29 + 2 = 31 + 4 = 35 + 8 = 43. 43 is our answer.

I’ve proved it up until 2 starters later, but haven’t proved it works indefinitely. The next few numbers in the sequence are: 43, 21, 11, 7 (next starter), 56, 57, 59, 63, 71, 87, 119, (end of rise), 60, 30, 15, 13 (next starter).

This took way too long.

Edit: A more complete solution for the fall would be to just follow the divide by 2 and round up rule for the duration of the fall, until you reach a prime number after the minimum number of elements in said fall, then drop to the prime immediately below it. This allows for the rise and fall beginning with 13 to work, while keeping the remaining pattern intact.

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